Chapter 18, Problem 73P

(a)

To determine

Find the equivalent resistance.

Answer to Problem 73P

The equivalent resistance is $35.0\Omega$.

Explanation of Solution

The sum of individual resistance connected in series and reciprocal of resistance connected in parallel will give the equivalent resistance.

From the given circuit, the resistor $R_2$, $R_3$  and $R_4$  are in parallel then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\ R\text{'}={\left(\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\right)}^{-1}\end{array}$

This resistance is in series with the $R_1$ and $R_5$ , then the final equivalent resistance is,

$R_{eq}=R_1+R\text{'}+R_5$

Substitute the value of $R\text{'}$ in the above equation.

$R_{eq}=R_1+{\left(\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\right)}^{-1}+R_5$

Conclusion:

Substitute, $15.0\Omega$ for $R_1$ , $40.0\Omega$ for $R_2$ , $20.0\Omega$ for $R_3$ , $40.0\Omega$ for $R_4$ ,and $10.0\Omega$ for $R_5$ in the above equation.

$\begin{array}{c}{R}_{eq}=15.0\Omega +{\left(\frac{1}{40.0\Omega }+\frac{1}{20.0\Omega }+\frac{1}{40.0\Omega }\right)}^{-1}+10.0\Omega \\ =35.0\Omega \end{array}$

Therefore, the equivalent resistance is $35.0\Omega$.

(b)

To determine

Find the current flows through $R_1$ .

Answer to Problem 73P

The current flows through $R_1$  is $0.686\text{ A}$.

Explanation of Solution

The emf is $24\text{ V}$.

Write the equation for current.

$I=\frac{\epsilon }{R_{eq}}$ (I)

Here, $R_{eq}$ is the equivalent resistance, $\epsilon$ is the emf and $I$ is the current.

Conclusion:

Substitute $24\text{ V}$ for $\epsilon$ and $35.0\Omega$ for $R_{eq}$ in equation I.

$\begin{array}{c}I=\frac{24\text{ V}}{35.0\Omega }\\ =0.686\text{ A}\end{array}$

Therefore, the current flows through $R_1$  is $0.686\text{ A}$.

(c)

To determine

Find the power dissipated in in the circuit.

Answer to Problem 73P

The power dissipated in the circuit is $16.5\text{ W}$.

Explanation of Solution

Write the equation for power dissipated.

$P=\frac{V^2}{R_{eq}}$ (II)

Here, $P$ is the power dissipated and $V$ is the potential difference.

Conclusion:

Substitute $\text{24}\text{.0 V}$ for $V$ and $35.0\Omega$ for $R_{eq}$ in equation II.

$\begin{array}{c}P=\frac{{\left(\text{24}\text{.0 V}\right)}^2}{35.0\Omega }\\ =16.5\text{ W}\end{array}$

Therefore, the power dissipated in the circuit is $16.5\text{ W}$.

(d)

To determine

Find the potential difference across $R_3$.

Answer to Problem 73P

The potential difference across $R_3$ is $6.9\text{ V}$.

Explanation of Solution

Use the Kirchhoff’s loop rule to the left hand loop to determine the voltage drop.

Consider $V_3$  as the potential difference across $R_3$

Conclusion:

From Kirchhoff’s loop rule,

$24.0\text{ V}-\left(0.686\text{ A}\right)\left(15.0\Omega \right)-V_3-\left(0.686\text{ A}\right)\left(10.0\Omega \right)=0$

From the above equation solve for $V_3$

$\begin{array}{c}{V}_2=\left(24.0-10.29-6.86\right)\text{ V}\\ =6.85\text{V}\\ =6.9\text{V}\end{array}$

Therefore, the potential difference across $R_3$ is $6.9\text{ V}$.

(e)

To determine

Find the current flows through $R_3$.

Answer to Problem 73P

The current flows through $R_3$ is $0.34\text{ A}$ .

Explanation of Solution

Write the equation for potential difference.

$I=\frac{V}{R}$ (III)

Here, $V$ is the potential difference, $R$ is the resistance and $I$ is the current.

Conclusion:

Substitute $6.85\text{V}$ for $V$ and $20.0\Omega$ for $R$ in equation III.

$\begin{array}{c}I=\frac{6.85\text{ V}}{20.0\Omega }\\ =0.34\text{ A}\end{array}$

Therefore, the current flows through $R_3$ is $0.34\text{ A}$.

(f)

To determine

Find the power dissipated in $R_3$.

Answer to Problem 73P

The power dissipated in $R_3$ is $2.4\text{ W}$ .

Explanation of Solution

Consider the resistor $R_3$ as $R_{eq}$

Conclusion:

Substitute $6.85\text{V}$ for $V$ and $20.0\Omega$ for $R_{eq}$ in equation II.

$\begin{array}{c}P=\frac{{\left(6.85\text{V}\right)}^2}{20.0\Omega }\\ =2.4\text{ W}\end{array}$

Therefore, the power dissipated in $R_3$ is $2.4\text{ W}$.